What is the magnitude of g at a height above earth's surface where free-fall acceleration equals 6.5m/s^2?

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about 

                     (1 / √0.66) = 1.23 times

the Earth's radius, so the height is about  910 miles above the surface.

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